using (5/2)x - (5/6)x^3 + ...
#3
Ok we can get sharper

f(x) = exp(x) -  exp(-x) + (7/10)(exp(- 5/4 x) - exp(- 9/4 x))

f(x) = (27 x)/10 - (49 x^2)/40 + (459 x^3)/320 - (2597 x^4)/3840 + (34329 x^5)/102400 + O(x^6)

(Taylor series)

This f(x) is periodic with 8 pi i, and bijective on R
and f'(0) = 2.7 what is close to e.


regards

tommy1729
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Messages In This Thread
using (5/2)x - (5/6)x^3 + ... - by tommy1729 - 04/25/2023, 11:25 PM
RE: using (5/2)x - (5/6)x^3 + ... - by tommy1729 - 04/26/2023, 12:27 PM
RE: using (5/2)x - (5/6)x^3 + ... - by tommy1729 - 04/26/2023, 11:29 PM
RE: using (5/2)x - (5/6)x^3 + ... - by tommy1729 - 04/29/2023, 11:00 PM



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